Swift 2 php mysql sign up issue - php

Question

I'm fairly new to programming with Swift in Xcode so I have been recently following the fantastic php sql sign up tutorial by Sergey Kargopolov located here http://swiftdeveloperblog.com/store-user-information-in-mysql-database/.
Unfortunately the tutorial was slightly outdated and needed to be updated for Swift 2.0. I managed to substitute the problematic code by implementing a do/catch statement which seems to work great except when i receive a response from the server side php script I can't seem to navigate to different view controllers based on the result. For example if the response is "Registration Successful" the user is directed to the protected page however if the email address already exists in the database, I want it to stay on the same page.
Currently I have the code working to the point where the alert appears (either user already exists or successful) but when the user clicks "Ok", they are directed to the protected page regardless of the result. I figured it would be a simple case of telling it to go to the next view based on the result in this "if" statement:
if (resultValue == "Success") {
isUserRegistered = true
let next = self.storyboard?.instantiateViewControllerWithIdentifier("ProtectedViewController") as! ProtectedViewController
self.presentViewController(next, animated: true, completion: nil)
}
but it doesn't seem to work. Hopefully it makes sense what i'm trying to do and any help is greatly appreciated.
here is my code:
let myUrl = NSURL(string: "http://localhost/PhpProject1/scripts/registerUser.php")
let request = NSMutableURLRequest(URL: myUrl!)
request.HTTPMethod = "POST"
let postString = "userEmail=\(userEmail!)&userFirstName=\(userFirstName)&userLastName=\(userLastName)&userPassword=\(userPassword!)"
request.HTTPBody = postString.dataUsingEncoding(NSUTF8StringEncoding);
let task = NSURLSession.sharedSession().dataTaskWithRequest(request) {
data, response, error in
if error != nil {
print("error=\(error)")
return
}
do {
let json = try NSJSONSerialization.JSONObjectWithData(data!, options: .MutableContainers) as? NSDictionary
if let parseJSON = json {
var resultValue = parseJSON["status"] as! String!
print("result: \(resultValue)")
var isUserRegistered: Bool = false
if (resultValue == "Success") {
isUserRegistered = true
}
var messageToDisplay: String = parseJSON["message"] as! String!
if (!isUserRegistered)
{
messageToDisplay = parseJSON["message"] as! String!
}
dispatch_async(dispatch_get_main_queue(), {
var myAlert = UIAlertController(title: "Alert", message: messageToDisplay, preferredStyle: UIAlertControllerStyle.Alert)
let okAction = UIAlertAction(title: "Ok", style: UIAlertActionStyle.Default) { action in
// self.dismissViewControllerAnimated(true, completion: nil)
let next = self.storyboard?.instantiateViewControllerWithIdentifier("ProtectedViewController") as! ProtectedViewController
self.presentViewController(next, animated: true, completion: nil)
}
myAlert.addAction(okAction)
}
)}
} catch { print(error)}
}
task.resume()
}
func displayAlertMessage(userMessage:String) {
let myAlert = UIAlertController(title: "Alert", message: userMessage, preferredStyle: UIAlertControllerStyle.Alert);
let okAction = UIAlertAction(title: "OK", style: UIAlertActionStyle.Default, handler: nil)
myAlert.addAction(okAction);
self.presentViewController(myAlert, animated: true, completion: nil)
}
}
I should also add that when the user registers successfully, I get "Result: 200" in the output window but when the user email already exists I get "Result: 400". I just don't know how to take advantage of this.
Thanks in advance!

Answer 1

From what I can see from your JSON response, parseJSON["status"] is the HTTP statusCode, which is a numerical status code. 200 being success, 400 being bad request, 404 being Not found, 500 being an error.
You could either read the message from the dictionary instead or depending on how your PHP code is working you could cast the status as an Integer and only segue when status is 200. For any other status code you likely want to show the message to the user.
This depends on whether you are intentionally returning status codes based on the result, for example a status code of 401 is un-authorised.
Using the status code properly would be much more reliable than comparing strings.
EDIT:
import Foundation
let jsonData = "{ \"message\": \"Successfully registered new user\", \"status\": 200, \"userEmail\": \"abc\", \"userFirstName\": \"a\", \"userId\": 22, \"userLastName\": \"a\" }".dataUsingEncoding(NSUTF8StringEncoding)
do {
let json = try NSJSONSerialization.JSONObjectWithData(jsonData!, options: .MutableContainers) as? NSDictionary
if let parseJSON = json {
let statusCode = parseJSON["status"] as! Int
switch statusCode {
case 200:
print("successfully logged in")
case 400, 401:
print("access denied")
default:
print("error occurred")
}
}
}
Above, I have added an example of how you might approach this in Swift, you just need to change the behaviour depending on which statusCodes you want to deal with and how

Answer 2

Well, I stayed up a little longer and thanks to you Scriptable I was able to figure it out.
Basically all I did was create a new variable:
var statuscheck: Int!
And then after the "if let parseJSON = json {" statement I cast the "status" as an Int like this:
self.statuscheck = (parseJSON["status"]! as! NSString).integerValue
Then after the line beneath the "dispatch_async(dispatch_get_main_queue()":
let okAction = UIAlertAction(title: "Ok", style: UIAlertActionStyle.Default) { action in
I did the check:
if (self.statuscheck == 200) {
let next = self.storyboard?.instantiateViewControllerWithIdentifier("ProtectedViewController") as! ProtectedViewController
self.presentViewController(next, animated: true, completion: nil)
}
This seems to work great now.
Thanks again for all your help Scriptable. You definitely pointed me in the right direction!